Question

If $m$ and $M$ are the minimum and the maximum values of $4 + \frac{1}{2} \sin^2 2x - 2\cos^4 x , x \in R$, then $M - m$ is equal to :

• A. $\frac{15}{4}$
• B. $\frac{9}{4}$
• C. $\frac{7}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$4+\frac{1}{2}sin^{2}2x-2 cos^{4}x$
$4+2\left(1-cos^{2}x\right)cos^{2}x-2 cos^{4}x$
$-4\left\{cos^{4}x-\frac{cos^{2}x}{2}-1+\frac{1}{16}-\frac{1}{16}\right\}$
$-4\left\{\left(cos^{2}x-\frac{1}{4}\right)^{2}-\frac{17}{16}\right\}$
$0\le cos^{2}x\le1$
$-\frac{1}{4}\le cos^{2}x-\frac{1}{4}\le \frac{3}{4}$
$0\le\left(cos^{2}x-\frac{1}{4}\right)^{2}\le\frac{9}{16}$
$-\frac{17}{16}\le\left(cos^{2}x-\frac{1}{4}\right)^{2}-\frac{17}{16}\le\frac{9}{16}-\frac{17}{16}$
$\frac{17}{14}\ge-4\left\{\left(cos^{2}x-\frac{1}{4}\right)^{2}-\frac{17}{16}\right\}\ge\frac{1}{2}$
$M=\frac{17}{4}$
$m=\frac{1}{2}$
$M-m=\frac{17}{4}-\frac{2}{4}=\frac{15}{4}$